Question: The cubic polynomial
\[8x^3 - 3x^2 - 3x - 1 = 0\]has a real root of the form $\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},$ where $a,$ $b,$ and $c$ are positive integers.  Find $a + b + c.$
Solution: We can arrange the equation as
\[9x^3 = x^3 + 3x^2 + 3x + 1 = (x + 1)^3.\]Taking the cube root of both sides, we get
\[x \sqrt[3]{9} = x + 1.\]Then $(\sqrt[3]{9} - 1)x = 1$, so
\[x = \frac{1}{\sqrt[3]{9} - 1}.\]To rationalize the denominator, we multiply the numerator and denominator by $\sqrt[3]{9^2} + \sqrt[3]{9} + 1.$  This gives us
\[\frac{\sqrt[3]{9^2} + \sqrt[3]{9} + 1}{(\sqrt[3]{9} - 1)(\sqrt[3]{9^2} + \sqrt[3]{9} + 1)} = \frac{\sqrt[3]{81} + \sqrt[3]{9} + 1}{8}.\]Then $a + b + c = 81 + 9 + 8 = \boxed{98}.$